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Probabilities about length of songs

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Hey!! :eek:

We have a playlist with $2000$ songs. The length of the songs on the playlist are on average $3.5$ minutes (i.e. $3$ minutes and $30$ seconds) with a standard deviation of $1.7$ minutes.

1) Can we find the probability that a randomly chosen song is longer than $4.5$ minutes?

2) Can we findthe probability that a random selection of $100$ songs lasts on average at least $4$ minutes?

3) Can we find the probability that a random selection of $200$ songs lasts in total at most $700$ minutes?




I have done the following:

Let $X_i$ be the RV that describes the length of the $i$-th song.

$\overline{X}_n=\frac{1}{n}(X_1+X_2+\ldots +X_n)$ is the mean.

$\overline{X}_n$ approximates, according the central limit theorem, a normal distribution with parameters $E(\overline{X}_n)=3.5$ and $V(\overline{X}_n)=\frac{\sigma_X^2}{n}=\frac{1.7^2}{n}=\frac{2.89}{n}$. Is this correct?

How could we find the probability at 1) ? (Wondering)


At 2) we want to calculate the probability $P(\overline{X}\geq 4)$ with $n=100$ right?

If this is correct, we have that $$P(\overline{X}\geq 4)=1-P(\overline{X}< 4)=1-\Phi \left (\frac{4-3.5}{\frac{1.7}{\sqrt{100}}}\right )\approx 1-\Phi (2.94)=1-0.9984=0.0016$$

Is everything correct? (Wondering)


At 3) we define a new RV $Z:=n\cdot \overline{X}_n$ which describes the sum of lengths of $n$ songs.

For $n=200$ we have that $E(Z)=E(200\overline{X}_{200})=200\cdot E(\overline{X}_{200})=200\cdot 3.5=700$ and $V(Z)=V(200\overline{X}_{200})=200^2\cdot V(\overline{X}_{200})=200^2\cdot \frac{2.89}{200}=578$

Then $$P(Z\geq 700)=\Phi \left (\frac{700-700}{\sqrt{578}}\right )=\Phi (0)=0.5$$

Is this correct? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,740
How could we find the probability at 1) ?
Hey mathmari !!

Doesn't one randomly chosen song have distribution $X\sim N(3.5, 1.7)$? (Wondering)
Then we need to find $P(X>4.5)$.

At 2) we want to calculate the probability $P(\overline{X}\geq 4)$ with $n=100$ right?

If this is correct, we have that $$P(\overline{X}\geq 4)=1-P(\overline{X}< 4)=1-\Phi \left (\frac{4-3.5}{\frac{1.7}{\sqrt{100}}}\right )\approx 1-\Phi (2.94)=1-0.9984=0.0016$$
Looks fine to me. (Nod)

At 3) we define a new RV $Z:=n\cdot \overline{X}_n$ which describes the sum of lengths of $n$ songs.

For $n=200$ we have that $E(Z)=E(200\overline{X}_{200})=200\cdot E(\overline{X}_{200})=200\cdot 3.5=700$ and $V(Z)=V(200\overline{X}_{200})=200^2\cdot V(\overline{X}_{200})=200^2\cdot \frac{2.89}{200}=578$

Then $$P(Z\geq 700)=\Phi \left (\frac{700-700}{\sqrt{578}}\right )=\Phi (0)=0.5$$
Shouldn't that be $P(Z \le 700)$? (Wondering)
Otherwise it looks fine to me.

Btw, we could also define $Z=X_1+...+X_{100}$, so that $EZ = 100\cdot EX_i$ and $\sigma^2(Z) = 100^2\cdot \sigma^2(X_i)$.
No need to introduce $\overline X$. (Nerd)